Solutions - Ncert

If the new length ( L' = 3L ) (three times original length), then to keep volume constant, the new area ( A' ) must satisfy:

[ R' = 9 \times 20 \ \Omega = 180 \ \Omega ] ncert solutions

New resistance: [ R' = \rho \frac{L'}{A'} = \rho \frac{3L}{A/3} ] [ R' = \rho \frac{3L \times 3}{A} = \rho \frac{9L}{A} ] [ R' = 9 \left( \rho \frac{L}{A} \right) ] [ R' = 9 \times R ] If the new length ( L' = 3L